Go through the proof very carefully. Make sure you understand all the steps of the proof, particularly the application of Euclidean division.

Statement

Let be a subgroup of , the group of integers under addition. Then, there are two possibilities:

is the trivial subgroup, i.e.

contains a smallest positive element, say , and is the set of multiples of . Thus, is an infinite cyclic group generated by , and is isomorphic to . We typically write .

Proof

Given: A nontrivial subgroup of , the group of integers under addition

To prove: There exists a smallest positive element in , and , so is isomorphic to

Proof: First, observe that if is nontrivial, then there exists a nonzero element in . This element may be either positive or negative. However, since is a subgroup, it is closed under taking additive inverses, so even if the element originally picked was negative, we have found a positive number in .

Thus, the set of positive numbers in is nonempty. Hence, there exists a smallest positive number in . Call it .

Clearly, all the integer multiples of are in . We need to prove that every element in is a multiple of .

By the Euclidean division algorithm, we can write:

where are integers and . Since , . Thus, is a nonnegative integer less than such that . By the minimality of , we have , so , as desired.

Thus, , or is the set of multiples of .

An explicit isomorphism from to is given by the map sending an integer to the integer .